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Alekssandra [29.7K]
2 years ago
9

PLEASE HELP FAST

Mathematics
2 answers:
mafiozo [28]2 years ago
7 0

Answer:

3b - 2 = 4

Step-by-step explanation:

3(2) = 6

6 - 2 = 4

luda_lava [24]2 years ago
3 0

Answer:

3b − 2 = 4

Step-by-step explanation:

2b + 24 = 30

3b − 2 = 4

b + 4 = 8

2b − 3 = 0

2b + 24 = 30

2(2) + 24 = 30

4 + 24 = 30

28 ≠ 30

-----------------------------------------------------------------------------------------------------------------

3b − 2 = 4

3(2) - 2 = 4

6 - 2 = 4

4 = 4

-----------------------------------------------------------------------------------------------------------------

b + 4 = 8

(2) + 4 = 8

6 ≠ 8

-----------------------------------------------------------------------------------------------------------------

2b - 3 = 0

2(2) - 3 = 0

4 - 3 = 0

1 ≠ 0

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When the expression $-2x^2-20x-53$ is written in the form $a(x+d)^2+e$, where $a$, $d$, and $e$ are constants, then what is the
Sladkaya [172]
We start with 2 x^{2} -20x-53 and wish to write it as a(x+d) ^{2} +e

First, pull 2 out from the first two terms: 2( x^{2} -10x)-53

Let’s look at what is in parenthesis. In the final form this needs to be a perfect square. Right now we have x^{2} -10x and we can obtain -10x by adding -5x and -5x. That is, we can build the following perfect square: x^{2} -10x+25=(x-5) ^{2}

The “problem” with what we just did is that we added to what was given. Let’s put the expression together. We have 2( x-5) ^{2}-53 and when we multiply that out it does not give us what we started with. It gives us 2 x^{2} -20x+50-53=2 x^{2} -20x-3

So you see our expression is not right. It should have a -53 but instead has a -3. So to correct it we need to subtract another 50.

We do this as follows: 2(x-5) ^{2}-53-50 which gives us the final expression we seek:

2(x-5) ^{2}-103

If you multiply this out you will get the exact expression we were given. This means that:
a = 2
d = -5
e =  -103

We are asked for the sum of a, d and e which is 2 + (-5) + (-103) = -106


7 0
3 years ago
Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie
ludmilkaskok [199]

Answer:

\lambda \geq 6.63835

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that X \sim Poisson(\lambda)

The probability mass function for the random variable is given by:

f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...

And f(x)=0 for other case.

For this distribution the expected value is the same parameter \lambda

E(X)=\mu =\lambda

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

P(X\geq 2)=1-P(X

Using the pmf we can find the individual probabilities like this:

P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}

P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}

And replacing we have this:

P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]

P(X\geq 2)=1-e^{-\lambda}(1+\lambda)

And we want this probability that at least of 99%, so we can set upt the following inequality:

P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99

And now we can solve for \lambda

0.01 \geq e^{-\lambda}(1+\lambda)

Applying natural log on both sides we have:

ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)

ln(0.01) \geq -\lambda+ln(1+\lambda)

\lambda-ln(1+\lambda)+ln(0.01) \geq 0

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}

Where :

f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)

f'(x_n)=1-\frac{1}{1+\lambda}

Iterating as shown on the figure attached we find a final solution given by:

\lambda \geq 6.63835

4 0
2 years ago
Solve x^2-3x=-8 PLEASE HELP ALGEBRA 2
Delvig [45]

Answer:

idk aljebra 2 either and i just got the same question how?

Step-by-step explanation:

5 0
2 years ago
which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater
Nutka1998 [239]

Answer:

\frac{\sqrt[3]{16y^4}}{x^2}

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices \sqrt[n]{a}  = a^{\frac{1}{n}}

So,

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} } gives

\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}

Also from laws of indices

(ab)^n = a^nb^n

So, the above expression can be further simplified to

\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}

Multiply the exponents gives

\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

Substitute 2^5 for 32

\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

From laws of indices

\frac{a^m}{a^n} = a^{m-n}

This law can be applied to the expression above;

\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})} becomes

2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}

Solve exponents

2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}

2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}

From laws of indices,

a^{-n} = \frac{1}{a^n}; So,

2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}} gives

\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}

The expression at the numerator can be combined to give

\frac{(2y)^{\frac{4}{3}}}{x^2}

Lastly, From laws of indices,

a^{\frac{m}{n} = \sqrt[n]{a^m}; So,

\frac{(2y)^{\frac{4}{3}}}{x^2} becomes

\frac{\sqrt[3]{(2y)}^{4}}{x^2}

\frac{\sqrt[3]{16y^4}}{x^2}

Hence,

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} } is equivalent to \frac{\sqrt[3]{16y^4}}{x^2}

8 0
3 years ago
If a car goes around a turn too quickly, it can leave tracks that form an arc of a circle. By finding the radius of the circle,
liubo4ka [24]
Given:
Segment AC = 130 feet
Segment CD = 70 feet

I think that I'll be using the Pythagorean Theorem in finding the value of r. r will be the hypotenuse

Segment CE = (r - 70 feet)

r² = a² + b²
r² = 130² + (r-70)²
r² = 16,900 + (r-70)(r-70)
r² = 16,900 + r² - 70r - 70r + 4900
r² - r² + 140r = 16,900 + 4,900
140r = 21,800
r = 21,800/140
r = 155.71 feet

The radius of the circle is 155.71 feet.

8 0
3 years ago
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